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ln (x)+ln (x-1)=ln (3x+12)

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(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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ln(x)+ln(x−1)=ln(3x+12)

Solution

x=6

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ln(x)+ln(x−1)=ln(3x+12)

Apply log rules

x(x−1)=3x+12

Solve x(x−1)=3x+12: x=6, x=−2

x=6, x=−2

Verify Solutions: x=6 True, x=−2 False

The solution is

x=6

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

ln (x)+ln (x-1)=ln (3x+12)

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